Specifically, the question says to consider the torus $T$ as a square with the usual identifications, with two opposite boundary edges labelled $a$ and the other two edges labelled $b$, and consider the wedge of two circles $S$ having two oriented loops labelled $\alpha$ and $\beta$. I have to show that there is no continuous map $T\to S$ which takes $a$ to $\alpha$ and $b$ to $\beta$ homeomorphically.
I should be able to tackle this with homology somehow, I figure by something like assuming there is such a map and showing that the homology groups don't agree, but I'm not seeing how to start going about it. The problem also tells me that it can also be solved with the fundamental group, but I'm even a little more lost as to how to attack that way. A nudge in the right direction would be much appreciated.
Well, the fundamental group of a torus is free abelian, the fundamental group of a wedge of circles is free. A continuous map induces a homomorphism on fundamental groups, and you know that YOUR map sends the generators of $\mathbb{Z} \times \mathbb{Z}$ to the generators of $F_2.$ But now consider the element $aba^{-1} b^{-1}$ it is identity in the abelian group, but its image is not the identity in the free group. Pretty goofy for a homomorphism.